Bạn tự giải đkxđ
\(\sqrt{x^2-4x+1}-\sqrt{9x^2+6x+1}=0\)
\(\Leftrightarrow\sqrt{x^2-4x+1}=\sqrt{9x^2+6x+1}\)
\(\Leftrightarrow x^2-4x+1=9x^2+6x+1\)
\(\Leftrightarrow8x^2-12x=0\Leftrightarrow4x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
Loại nghiệm 3/2 vì k thỏa mãn đk nhé.
\(\sqrt{x^2-4x+4}-\sqrt{9x^2+6x+1}=0\)
xin lỗi mình nhầm
\(\sqrt{x^2-4x+4}-\sqrt{9x^2+6x+1}=\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(3x+1\right)^2}\)
\(\Leftrightarrow\left|x-2\right|=\left|3x+1\right|\)
\(\Leftrightarrow\pm\left(x-2\right)=3x+1\)
Nếu \(x-2=3x+1\)
\(\Leftrightarrow-2x=3\)
\(\Leftrightarrow x=-\frac{3}{2}\)
Nếu \(-\left(x-2\right)=3x+1\)
\(\Leftrightarrow2-x=3x+1\)
\(\Leftrightarrow4x=1\)
\(\Leftrightarrow x=\frac{1}{4}\)
